Prove the following for negative real numbers: a < b implies a² > b²?

Question

Prove the following for all negative real numbers:

If a < b, it can be implied that a² > b².

Here's my proof, please check.

If the numbers are negative, the lesser it is, the higher its absolute value. So if a < b, then |a| > |b|.

Square any number and it becomes positive, so we can say that a² = |a|² and b² = |b|².

|a|² > |a| * |b|

|a| * |b| > |b|²

By transitive law, |b|² < |a|²

Since |a|² = a², and |b|² = b²...

b² < a²

a² > b²

Q.E.D.

Please let me know if I proved it wrongly, or if there's a better way to prove it. :) Thank you.

Answer 1

Your proof is all right. However you don&#039;t need to mention the following relations in the proof

|a|² &gt; |a| * |b|

|a| * |b| &gt; |b|²

PROOF:

a&lt;0 and b&lt;0:

a&lt;b==&gt;|a|&gt;|b|==&gt;|a|²&gt;|b|²

==&gt;a²&gt;b²

Answer 2

The proof seems correct.

Good job realizing that you need to use absolute value.

Answer 3

Your proof looks ok, but here&#039;s one that doesn&#039;t involve absolute values:

a&lt;b =&gt; b-a&gt;0 ...(1)

since a and b are both negative a+b is also negative, meaning

b+a&lt;0 =&gt; -(b+a)&gt;0 ...(2)

Now positive * positive still positive, hence (1)*(2) give us:

-(b-a)(b+a)&gt;0

=&gt; -(b²-a²)&gt;0

=&gt; -b²+a²&gt;0

=&gt; a²&gt;b²

Answer 4

In my opinion,your proof is right. I think, you need not use the

concept of absolute value. My suggestion is:

if a&lt;b (both a,b are -ve),then

a^2&gt;ab &amp; ab&gt;b^2=&gt;

a^2&gt;b^2

Q.E.D.